Three dice are tossed. What is the probability that the numbers shown will all be different?
Probability is the branch of mathematics that deals with the possibility of occurrence of an event among all other possibilities. It has many applications in the field of electronics and electrical engineering, as well as in advanced mathematical studies.
Answer: When three dice are tossed, 5/9 is the probability that the numbers shown will all be different.
Let's understand the solution step by step.
Now, we are given 3 fair dice. Each of the dice has 6 sides.
So, the total number of possible outcomes = 6 × 6 × 6 = 216
Now, if none of the numbers has to be repeated, then let's follow the below steps to find the number of favourable outcomes:
- Initially, the first die has all the 6 possibilities. Now, let's assume it shows up 5. Now 5 can't appear on the other die. Hence, the number of outcomes can be 5.
- Now, let's assume the second die shows up number 4. Now, 4 and 5 can't be repeated. Hence, the number of outcomes now is 4.
- Now, the third die can show up any of the four numbers left.
Therefore, the total number of favourable outcomes in which all the numbers are different = 6 × 5 × 4 = 120.
The probability required = favourable outcomes / All possible outcomes = 120 / 216 = 5/9.
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